The three distinct operators C++ uses to access the members of a class or class object, namely the double colon ::, the dot ., and the arrow ->, are used for three different scenarios that are always well-defined. Knowing this allows you to immediately know quite a lot about a and b just by looking at a::b, a.b, or a->b, respectively, in any code you look at.
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a::bis only used ifbis a member of the class (or namespace)a. That is, in this caseawill always be the name of a class (or namespace). -
a.bis only used ifbis a member of the object (or reference to an object)a. So fora.b,awill always be an actual object (or a reference to an object) of a class. -
a->bis, originally, a shorthand notation for(*a).b. However,->is the only of the member access operators that can be overloaded, so ifais an object of a class that overloadsoperator->(common such types are smart pointers and iterators), then the meaning is whatever the class designer implemented. To conclude: Witha->b, ifais a pointer,bwill be a member of the object the pointerarefers to. If, however,ais an object of a class that overloads this operator, then the overloaded operator functionoperator->()gets invoked.
The small print:
- In C++, types declared as
class,struct, orunionare considered “of class type”. So the above refers to all three of them. - References are, semantically, aliases to objects, so I should have added “or reference to a pointer” to the #3 as well. However, I thought this would be more confusing than helpful, since references to pointers (
T*&) are rarely ever used. - The dot and arrow operators can be used to refer to static class members from an object, even though they are not members of the object. (Thanks to Oli for pointing this out!)